Dual port ram

I came across the below code for dual port ram.
what does rd_data_2[(i+1)minWIDTH-1:iminWIDTH] <= RAM[{addr_2, lsbaddr}]** mean in this code?
is there any bug in this code?

module ram_dual_port#(
    parameter WIDTH       = 16,
    parameter DEPTH       = 1024,
    parameter ADDRWIDTHA  = $clog2(DEPTH),
    parameter WIDTHB      = WIDTH,
    parameter DEPTHB      = DEPTH,
    parameter ADDRWIDTHB  = $clog2(DEPTHB)
) (
    input                         clk1,
    input                         clk2,
    input                         wr_valid_1,
    input                         wr_valid_2,
    input       [ADDRWIDTHA-1:0]  addr_1,
    input       [ADDRWIDTHB-1:0]  addr_2,
    input       [WIDTH-1:0]       wr_data_1,
    input       [WIDTHB-1:0]      wr_data_2,
    output reg  [WIDTH-1:0]       rd_data_1,
    output reg  [WIDTHB-1:0]      rd_data_2
);
    
    `define max(a,b) {(a) > (b) ? (a) : (b)}
    `define min(a,b) {(a) < (b) ? (a) : (b)}
    
    function integer log2;
        input integer value;
        reg [31:0] shifted;
        integer res;
    begin
        if (value < 2)
            log2 = value;
        else
        begin
            shifted = value-1;
            for (res=0; shifted>0; res=res+1)
                shifted = shifted>>1;
            log2 = res;
        end
    end
    endfunction
    
    localparam maxSIZE   = `max(DEPTH, DEPTHB);
    localparam maxWIDTH  = `max(WIDTH, WIDTHB);
    localparam minWIDTH  = `min(WIDTH, WIDTHB);
    localparam RATIO     = maxWIDTH / minWIDTH;
    localparam log2RATIO = log2(RATIO);
    
    reg     [minWIDTH-1:0]  RAM [0:maxSIZE-1];
    
    reg     [WIDTHB-1:0]  readB;
    
    genvar i;
    
    always @(posedge clk1)
    begin
        if (wr_valid_1)
            RAM[addr_1] <=  wr_data_1;
        rd_data_1       <=  RAM[addr_1];
    end
    
    generate for (i = 0; i < RATIO; i = i+1)
        begin: ramwrite
            localparam [log2RATIO-1:0] lsbaddr = i;
            always @(posedge clk2)
            begin
                rd_data_2[(i+1)*minWIDTH-1:i*minWIDTH]  <=  RAM[{addr_2, lsbaddr}];
            end
        end
    endgenerate

endmodule

Say WIDTH = 8 (a byte), and WIDTHB is some multiple of a byte, eg 4*8.

The number of bytes of your RAM will be 1.

The number of bytes of 2nd port (the one which width is set by WIDTHB parameter) will be equal to your RATIO = 4.

You are generating a clocked always block which will model an output data register for the 2nd port. Its width will also be equal to RATIO (4) bytes, while your RAM width is 1 byte.

What this statement does is basically: for each byte-wide part of the whole rd_data_2, assign (on posedge clk) a byte of RAM array, with address given by the input address and a position of this byte-wide part inside rd_data_2.

For example if rd_data_2 is 32 byte wide then it has four byte-wide parts, with positions 0, 1, 2 and 3. This will make your ā€œiā€.

say rd_data_2 = 0xFECE1100

00 will be position 0
11 will be position 1
CE will be position 2
FE will be position 3

The addr_2 will address a part of RAM array which is the same width as rd_data_2 - four byte wide. However, the RAM array is 1 byte wide. That means it will have two more bits of address, and you will need those bits to pull out the data.

Therefore to assign 4 consecutive bytes from 4 consecutive addresses to appropriate rd_data_2 positions you must concatanate input address and byte position number. That is to fill the missing 2 address bits.

Say your content is:
ADDR DATA
0011 CE
0100 FE
0101 FE
0101 FE
0111 FE
1000 CE

And your output will be:
FEFEFEFE

As you can see to address the 4 bytes with FE content at once you only need the two MSBs, as they are the same. However, to pull data out of RAM you need to supply the two LSBs. That is what the concatanation is for.

hii isthere any code for dual port ram using synchronous and asynchronous using system verilog or verilog

In reply to mohanbabu Babu Bollineni:

hi, is there any code for dual port ram using synchronous and asynchronous using system verilog or verilog

Sure, there are plenty. https://lmgtfy.com/?q=code+for+dual+port+ram+using+synchronous+and+asynchronous+using+system+verilog+or+verilog